Weak convergence on $L^\infty([0,1])$

Question

It is well known that the dual space of $L^\infty([0,1])$ is pretty large and I do not really have a feeling for it to be honest. Currently, I am interested in the following question:

Suppose that you have a bounded sequence $(f_n)_n$ in $L^\infty([0,1])$ that converges almost everywhere to some function $f \in L^\infty([0,1])$. Does it follow that $f_n \to f$ weakly?

I know that $\langle f_n, g \rangle \to \langle f, g \rangle$ for each $L^1([0,1])$ by Lebesgue's dominated convergence theorem but that suffice to show that $f_n \to f$ weakly? In particular, is $L^1$ dense in the dual space of $L^\infty$? If so, does the result still hold in for the Bochner space $L^\infty([0,1]; X)$, where $X$ is some Banach space? It would be also nice to see some reference, if the answer to that question is well known.

Answer 1

No. (But of course this does not work in ZF, it requires Hahn-Banach.)

We may consider the subspace $C[0,1]$ of continuous functions, $C[0,1] \subset L^\infty[0,1]$. The usual sup norm on $C[0,1]$ agrees with the norm obtained by restricting the norm on $L^\infty[0,1]$.
Here is a linear functional $\phi$ on $C[0,1]$: $$ \phi(f) = f(0)\qquad\text{for all } f \in C[0,1]. $$ This functional has norm $1$ as a functional on $C[0,1]$.

By the Hahn-Banach theorem, there is an extension: $\Phi$ is a linear functional on $L^\infty[0,1]$, $\Phi(f) = \phi(f)$ if $f \in C[0,1]$, and $\Phi$ has norm $1$.

Now consider the sequence $f_n \in L^\infty[0,1]$ $$ f_n(x) = \begin{cases}1-nx,\quad & 0\le x \le \frac{1}{n},\\ 0,\qquad & \frac{1}{n} < x \le 1.\end{cases} $$ Also consider the constant function $f(x) = 0$ for all $x$.
Note that $f_n, f$ are in $L^\infty[0,1]$ and $f_n(x) \to f(x)$ for almost all $x \in [0,1]$.
But also $f_n, f \in C[0,1]$ so $\Phi(f_n) = \phi(f_n) = 1$ for all $n$ while $\Phi(f) = \phi(f) = 0$. So $f_n$ does not converge weakly to $f$.